If the vertices of a hyperbola be at (- 2, 0) and (2, 0) and one of...

Since, the vertex of the hyperbola are ${A}^{′}$\left(- 2, 0\right)$and$A\left(2, 0\right).

Hence, the transverse axis of the hyperbola is$x -$ axis and centre is $O\left(0, 0\right)$.

Equation of hyperbola can be taken as

$\dfrac{{x}^{2}}{{a}^{2}} - \dfrac{{y}^{2}}{{b}^{2}} = 1$

$⇒ a = 2 and ae = 3$

$⇒ e = \dfrac{3}{2}$

Since, ${b}^{2} = {a}^{2}\left({e}^{2} - 1\right) ⇒ {b}^{2} = {a}^{2}{e}^{2} - {a}^{2} = 9 - 4 = 5$

Hence, equation of hyperbola is $\dfrac{{x}^{2}}{4} - \dfrac{{y}^{2}}{5} = 1$

Clearly, point \left(6, 5\sqrt{2}\right) doesn't lies on the hyperbola.